# A spring with a constant of 6/5 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 7/5 kg and speed of 5/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Feb 17, 2016

$\approx 1.350 \text{m}$

#### Explanation:

As you have already noted, this question is about the conservation of energy.

All of the kinetic energy of the object will turn into elastic potential energy in the spring.

Since a spring constant with units ${\text{kg/s}}^{2}$ is given, it is assumed that the spring obeys Hooke's Law.

This means that the force required to stretch/compress the spring is proportional to how far the spring is already stretched/compressed.

Mathematically, we write,

${F}_{\text{spring}} = - k x$

The negative sign arises from the fact that the springs always wants to restore itself to its equilibrium length, and therefore, the force will always oppose the spring's displacement.

The potential energy of such a spring, $U$, can be found by integration.

$U = - \int {F}_{\text{spring}} \cdot \mathrm{dx} = \frac{1}{2} k {x}^{2}$

So initially, the kinetic energy of the object , $\text{KE}$, is

$\text{KE} = \frac{1}{2} m {v}^{2}$

$= \frac{1}{2} \cdot {\left(\frac{7}{5} \text{kg")*(5/4"m/s}\right)}^{2}$

$= \frac{35}{32} \text{J}$

This means that the amount of energy stored in the spring in the end is $\frac{35}{32} \text{J}$.

$\frac{1}{2} k {x}^{2} = \frac{35}{32} \text{J}$

1/2 (6/5"kgs"^{-2}) x^2 = 35/32 "J"

$x = \pm 1.350 \text{m}$